a.5/14+1/2=6/7

b.5/3-1=2/3

c.4/3-1/3=1

a) 5/14+ 1/2= 6/7

b) 5/3 - 1= 2/3

c) 4/3- 1/3 =1

a.5/14+1/2=6/7

b.5/3-1=2/3

c.4/3-1/3=1

`1`

`a, 1/2 +1/3= 3/6 + 2/6 =5/6`

`d, 1/3 +3/5= 5/15 + 9/15=14/15`

`c,4/5 +1/2= 8/10 + 5/10= 13/10`

`2`

`a,1/2 +1/4=2/4 +1/4=3/4`

`b, 2/3 +1/6 = 4/6+1/6=5/6`

`c, 7/12 +1/2=7/12+ 6/12= 13/12`

`3`

Giải

Cả `2` ngày đi tất cả số quãng đường là :

`1/4 +1/2 =1/4+ 2/4= 3/4 ( quãng đường)`

đ/s...

`@ yL`

Ta co:\(B=\frac{2008}{1}+\frac{2007}{2}+...+\frac{2}{2007}+\frac{1}{2008}\)

           \(B=\frac{2009-1}{1}+\frac{2009-2}{2}+...+\frac{2009-2007}{2007}+\frac{2009-2008}{2008}\)

            \(B=\left(\frac{2009}{1}+\frac{2009}{2}+...+\frac{2009}{2008}\right)-\left(\frac{1}{1}+\frac{2}{2}+...+\frac{2008}{2008}\right)\)

            \(B=2009+2009\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2008}\right)-2008\)

            \(B=1+2009\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2008}\right)\)

             \(B=2009\left(\frac{1}{2}+\frac{1}{3}+....+\frac{1}{2008}+\frac{1}{2009}\right)\)

Vay \(\frac{A}{B}=\frac{1}{2009}\)

mik đọc nhầm đề rồi.Kết quả là 9/187

Li-ke cho mik nhé!

\(a,\dfrac{1}{2-\sqrt{3}}-3\sqrt{\dfrac{1}{3}}+\sqrt{12}\\ =\dfrac{2+\sqrt{3}}{\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)}-\dfrac{\sqrt{3^2}}{\sqrt{3}}+\sqrt{2^2.3}\\ =\dfrac{2+\sqrt{3}}{4-3}-\sqrt{3}+2\sqrt{3}\\ =2+\sqrt{3}-\sqrt{3}+2\sqrt{3}\\ =2+2\sqrt{3}\)

\(b,\dfrac{2}{1+\sqrt{2}}-\sqrt{9-\sqrt{32}}\\ =\dfrac{2\left(1-\sqrt{2}\right)}{\left(1+\sqrt{2}\right)\left(1-\sqrt{2}\right)}-\sqrt{9-4\sqrt{2}}\\ =\dfrac{2-2\sqrt{2}}{1-2}-\sqrt{\left(2\sqrt{2}\right)^2-2.2\sqrt{2}+1}\\ =-2+2\sqrt{2}-\sqrt{\left(2\sqrt{2}-1\right)^2}\\ =-2+2\sqrt{2}-\left|2\sqrt{2}-1\right|\\ =-2+2\sqrt{2}-2\sqrt{2}+1\\ =-1\)

- Đặt \(A=1-\frac{1}{2^2}-\frac{1}{3^2}-...-\frac{1}{2006^2}\)

- Ta có: \(1=1\)

          \(\frac{1}{2^2}>\frac{1}{2.3}\)

          \(\frac{1}{3^2}>\frac{1}{3.4}\)

          \(................\)

   \(\frac{1}{2006^2}>\frac{1}{2006.2007}\)

 \(\Rightarrow A>1-\frac{1}{2.3}-\frac{1}{3.4}-\frac{1}{4.5}-...-\frac{1}{2006.2007}\)

\(\Leftrightarrow A>1-\left(\frac{1}{2}-\frac{1}{3}\right)-\left(\frac{1}{3}-\frac{1}{4}\right)-...-\left(\frac{1}{2006}-\frac{1}{2007}\right)\)

\(\Leftrightarrow A>1-\frac{1}{2}+\frac{1}{3}-\frac{1}{3}+\frac{1}{4}-...-\frac{1}{2006}+\frac{1}{2007}\)

\(\Leftrightarrow A>1+\frac{1}{2007}=\frac{2008}{2007}\)mà \(\frac{2008}{2007}>1>\frac{1}{2006}\)

 \(\Rightarrow A>\frac{1}{2006} \left(ĐPCM\right)\)

^_^ Chúc bạn hok tốt ^_^

Bài 1: 

a: \(=\dfrac{4}{8}+\dfrac{2}{8}+\dfrac{1}{8}=\dfrac{7}{8}\)

b: \(=\dfrac{4}{12}+\dfrac{2}{12}+\dfrac{1}{12}=\dfrac{7}{12}\)

B1:

a \(\dfrac{7}{8}\)

b \(\dfrac{7}{12}\)

B2:

a \(\dfrac{3}{7}\)

b \(\dfrac{5}{9}\)

c \(\dfrac{6}{5}\)

B3:

 đội còn phải sửa phần quãng đường nữa  :\(\dfrac{8}{15}\) phần quãng đường

a: =20/60+12/60-15/60=17/60

b: =7/8-1/4-2/5=5/8-2/5=25/40-16/40=9/40

\(a,\dfrac{1}{3}+\dfrac{1}{5}-\dfrac{1}{4}=\dfrac{20}{60}+\dfrac{12}{60}-\dfrac{15}{60}=\dfrac{17}{60}\\ b,\dfrac{7}{8}-\left(\dfrac{1}{4}+\dfrac{2}{5}\right)=\dfrac{7}{8}-\dfrac{13}{20}=\dfrac{9}{40}\)